Circles — Semicircle & Radii

Section 1 of 3

Angle in a Semicircle

Standing on the diameter

An angle in a semicircle (standing on the diameter) is always $90^\circ$.

Worked example — find the third angle

A triangle sits in a semicircle with a $90^\circ$ angle and a $30^\circ$ angle. Find $A$.

Angle in a semicircle $= 90^\circ$

$A + 90 + 30 = 180 \Rightarrow A = 60^\circ$

$A = 60^\circ$

You try

A triangle in a semicircle has the $90^\circ$ angle and two equal angles $A$. Find $A$.

$A + A + 90 = 180$.

$2A = 90 \Rightarrow A = 45^\circ$

$A = 45^\circ$

Section 2 of 3

Two Radii Make an Isosceles Triangle

Equal radii

Two radii of the same circle are equal, so they form an isosceles triangle with equal base angles.

Worked example — apex angle given

Two radii make a triangle with apex angle $140^\circ$ at the centre and base angles $A$. Find $A$.

Base angles equal: $A + A + 140 = 180$

$2A = 40 \Rightarrow A = 20^\circ$

$A = 20^\circ$

You try

Two radii form an isosceles triangle with a base angle of $55^\circ$. Find the apex angle $A$ at the centre.

$A + 55 + 55 = 180$.

$A = 180 - 110 = 70^\circ$

$A = 70^\circ$

Section 3 of 3

Finding the Radius

Worked example — right angle then Pythagoras

A right-angled triangle sits in a circle with the right angle on the circle (angle in a semicircle). Its short sides are $8$ and $6$. The hypotenuse is the diameter $H$. Find the radius.

$H^2 = 8^2 + 6^2 = 64 + 36 = 100 \Rightarrow H = 10$

$H$ is the diameter, so $r = \tfrac{10}{2} = 5$

$r = 5$

You try

A circle has radius $6$, so the diameter $|AB| = 12$. A point $C$ on the circle makes a right angle (angle in a semicircle). If $|AB| = 12$ and $|AB|$ is the hypotenuse with one side $8$, find $|AC|$.

Right angle at $C$: $|AC|^2 + 8^2 = 12^2$.

$|AC|^2 = 144 - 64 = 80 \Rightarrow |AC| = \sqrt{80}$

$|AC| = \sqrt{80}$

That’s Class 8.

Angle in a semicircle ($90^\circ$), two radii (isosceles), and finding a radius. Class 9: enlargements.